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This paper deals with the existence of triple positive solutions for a type of second-order
singular boundary problems with general differential operators. By using the Leggett-Williams
fixed point theorem, we establish an existence criterion for at least three positive
solutions with suitable growth conditions imposed on the nonlinear term.

1. Introduction

In this paper, we study the existence of triple positive solutions for the following
second-order singular boundary value problems with general differential operators:

(11)

where , , and with

(12)

It is easy to see that and may be singular at and/or

When or and , the two kinds of singular boundary value problems have been discussed extensively
in the literature; see [1–10] and the references therein. Hence, the problem that we consider is more general
and is different from those in previous work.

Furthermore, we will see in the later that the presence of brings us three main difficulties:

(1)the Green's function cannot be explicitly expressed;

(2)the equivalence between BVP (1.1) and its associated integral equation has to be
proved;

(3)the compactness of associated integral operator has to be verified.

We will overcome the above mentioned difficulties in Section 2. Also, although the
Leggett-William fixed point theorem is used extensively in the study of triple positive
differential equations, the method has not been used to study this type of second-order
singular boundary value problem with general differential operators. We are concerned
with solving these problems in this paper.

To state our main tool used in this paper, we give some definitions and notations.

Let be a real Banach space with a cone . A map is said to be a nonnegative continuous concave functional on if is a continuous and

(13)

for all and . Let be two numbers such that and a nonnegative continuous concave functional on . We define the following convex sets:

(14)

Theorem 1.1 (Leggett-Williams fixed point theorem).

Let be completely continuous, and let be a nonnegative continuous concave functional on such that for all . Suppose that there exist such that

(i) and , for ;

(ii), for ;

(iii), for with .

Then has at least three fixed points in satisfying and .

Remark 1.2.

We note the existence of triple positive solutions of other kind of boundary value
problems; see He and Ge [11], Zhao et al. [12], Zhang and Liu [13], Graef et al. [14], and the references therein.

The rest of the paper is organized as follows. In Section 2, we overcome the above-mentioned
difficulties in this work. The main results are formulated and proved in Section 3.
Finally, an example is presented to demonstrate the application of the main theorems
in Section 4.

2. Preliminaries and Lemmas

Throughout this paper, we assume the following:

(H1);

(H2), and ;

(H3) is continuous and does not vanish identically on any subinterval of , and ;

(H4) is continuous.

Lemma 2.1.

Suppose that (H1) and (H2) hold. Then

(i)the initial value problem

(21)

has a unique solution and ;

(ii)the initial value problem

(22)

has a unique solution and .

Proof.

We only prove (i). (ii) can be treated in the same way.

Suppose that and is a solution of (2.1), that is,

(23)

Let

(24)

Multiplying both sides of (2.3) by , then

(25)

Since and , integrating (2.5) on , we have

(26)

Moreover, integrating (2.6) on , , we have

(27)

Let

(28)

Clearly, , and (2.7) reduces to

(29)

By using Fubini's theorem, we have

(210)

Therefore,

(211)

which implies that is a solution of integral equation (2.11).

Conversely, if is a solution of (2.11) with , by reversing the above argument we could deduce that the function is a solution of (2.1) and satisfy and . Therefore, to prove that (2.1) has a unique solution, , and is equivalent to prove that (2.11) has a unique solution .

To do this, we endow the following norm in :

(212)

Let be operator defined by

(213)

Since

(214)

then, is well defined. Set

(215)

Then, for any ,

(216)

and subsequently,

(217)

Thus,

(218)

Since , has a unique fixed point by Banach contraction principle. That is, (2.11) has a unique solution .

From this fact, it is easy to verify that is equicontinuous. Therefore, by the Arzela-Ascoli theorem, is a completely continuous operator.

3. Main Result

Let be nonnegative continuous concave functional defined by

(31)

We notice that, for each , , and also that by Lemma 2.6, is a solution of (1.1) if and only if is a fixed point of the operator .

For convenience we introduce the following notations. Let

(32)

Theorem 3.1.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(S1), for ;

(S2), for ;

(S3), for .

Then the boundary value problem (1.1) has at least three positive solutions in satisfying , and .

Proof.

From Lemma 2.9, is a completely continuous operator. If , then , and assumption (S3) implies that . Therefore

(33)

Hence, . In the same way, if , then . Therefore, condition (ii) of Leggett-williams fixed-point theorem holds.

To check condition (i) of Leggett-Williams fixed-point theorem, choose . It is easy to see that and . so,

(34)

Hence, if , then . From assumption (S2) and Remark 2.8, we have

(35)

Finally, we assert that if and , then . To see this, suppose that and , then

(36)

To sum up, all the conditions of Leggett-williams fixed-point theorem are satisfied.
Therefore, has at least three fixed points, that is, problem (1.1) has at least three positive
solutions in satisfying and .

Theorem 3.2.

Assume that (H1)–(H4) hold. Let , and suppose that satisfies the following conditions:

(A1), for ;

(A2), for .

Then the boundary value problem (1.1) has at least positive solutions.

Proof.

When , it follows from condition (A1) that , which means that has at least one fixed point by the Schauder fixed-point Theorem. When , it is clear that Theorem 3.1 holds (with ). Then we can obtain at least three positive solutions , and satisfying and with . Following this way, we finish the proof by the induction method.

4. Example

Consider the following boundary value problem:

(41)

where

(42)

Then, by computation, we have

(43)

Furthermore, for ,

(44)

In fact, let , then , and . It is easy to compute that

(45)

Then, , that is

(46)

The other inequalities in (4.4) can be proved by the same method.

Thus, we can choose that , and . By computation, we have

(47)

Let , and . Then, we can compute

(48)

Consequently,

(49)

Therefore, all the conditions of Theorem 3.1 are satisfied, then problem (4.1) has
at least three positive solutions , and satisfying